Hello,

I have some questions about my homework. I need to prove the following limits of sequences using the limit definition:

1. $\displaystyle \lim_{n\to\infty} \frac{2n^2-6n+1}{3n^2+n-3} = \frac{2}{3}$

2. $\displaystyle \lim_{n\to\infty} (\frac{2}{3})^n = 0$

my proofs look like this:

1. for every $\displaystyle \varepsilon > 0$ we will choose $\displaystyle N(\varepsilon) = 2 + \frac{3}{\varepsilon}$, and then:

$\displaystyle \forall n > N(\varepsilon) \left | \frac{2n^2-6n+1}{3n^2+n-3} - \frac{2}{3} \right | = \left | \frac{-20n+9}{9n^2+3n-9} \right | \underset{\forall n\geq 1}{=} \frac{20n-9}{9n^2+3n-9} < \frac{20n+4n}{9n^2+(3n-9)} \underset{\forall n\geq 2}{<} \frac{24n}{9n^2-n^2} = \frac{3}{n} < \frac{3}{N} = \frac{3}{2 + \frac{3}{\varepsilon}} < \frac{3}{\frac{3}{\varepsilon}}= \varepsilon$

2. for every $\displaystyle \varepsilon > 0$ we will choose $\displaystyle N(\varepsilon) = \log_{\frac{2}{3}}\varepsilon$, and then:

$\displaystyle \forall n > N(\varepsilon) \left | (\frac{2}{3})^n - 0 \right | = (\frac{2}{3})^n \underset{\forall n > N}{<} (\frac{2}{3})^N = (\frac{2}{3})^{log_{\frac{2}{3}}\varepsilon} = \varepsilon$

I'm new to this kind of proofs and I'm not sure if this correct,

anyone can tell me if it's the correct way to do this?

Thank you.